Algebra Expressions MCQ Questions

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Algebra Expressions MCQ Questions with details Solution

 

Q11. If p = 999, then the value of $\sqrt[3]{\mathrm{p}\left( \mathrm{p}^2+3\mathrm{p}+3 \right) +1}$is

(a) 998

(b) 999

(c) 1000

(d) 1002

Answer: (c) 1000

Solution: $\sqrt[3]{\mathrm{p}\left( \mathrm{p}^2+3\mathrm{p}+3 \right) +1}\,\,$

$=\,\,\sqrt[3]{\mathrm{p}^3+3\mathrm{p}^2.1+3\mathrm{p}.1^2+1}$

$=\,\,\sqrt[3]{\left( \mathrm{p}+1 \right) ^3}$

= p + 1 = 999 + 1 = 1000

Q12. If x + $\frac{1}{\mathrm{y}}$= 1 and y + $\frac{1}{\mathrm{z}}$ = 1, what is the value of xyz?

(a) 1

(b) – 1

(c) 0

(d) ½

Answer: (b) – 1

Solution: y + $\frac{1}{\mathrm{z}}$ = 1 ⇒ y = 1 – $\frac{1}{\mathrm{z}}$= $\frac{\mathrm{z}-1}{\mathrm{z}}$

Now x + $\frac{1}{\mathrm{y}}$= 1

⇒ xy + 1 = y

⇒ xy + 1 = $\frac{\mathrm{z}-1}{\mathrm{z}}$

⇒ xyz + z = z –1

⇒ xyz = –1

Q13. If x2 + y2 + 2x + 1 = 0, then the value of x31 + y35 is

(a) – 1

(b) 0

(c) 1

(d) 2

Answer: (a) – 1

Solution: x2 + y2 + 2x + 1 = 0

⇒ x2 + 2x + 1 + y2 = 0

⇒ (x + 1)2 + y2 = 0

⇒ x + 1 = 0 and y = 0

⇒ x – 1 and y = 0

∴ x31 + y35 = – 1 + 0 = –1

Q14. If x2 + y2 + 2x + 1 = 0, then the value of x2020 + y2020 is

(a) – 1

(b) 0

(c) 1

(d) 2

Answer: (c) 1

Solution: x2 + y2 + 2x + 1 = 0

⇒ x2 + 2x + 1 + y2 = 0

⇒ (x + 1)2 + y2 = 0

⇒ x + 1 = 0 and y = 0

⇒ x – 1 and y = 0

∴ x2020 + y2020 =  1 + 0 = 1

Q15.  $\frac{\mathrm{a}}{1-\mathrm{a}}+\frac{\mathrm{b}}{1-\mathrm{b}}+\frac{\mathrm{c}}{1-\mathrm{c}}=1$ then find the value of

$\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1$

(a) 1

(b) 2

(c) 3

(d) 4

Answer: (a) 1

Solution: $\frac{\mathrm{a}}{1-\mathrm{a}}+\frac{\mathrm{b}}{1-\mathrm{b}}+\frac{\mathrm{c}}{1-\mathrm{c}}=1$

⇒$\frac{\mathrm{a}}{1-\mathrm{a}}+1+\frac{\mathrm{b}}{1-\mathrm{b}}+1+\frac{\mathrm{c}}{1-\mathrm{c}}+1=4$

⇒$\frac{\mathrm{a}+1-\mathrm{a}}{1-\mathrm{a}}+\frac{\mathrm{b}+1-\mathrm{b}}{1-\mathrm{b}}+\frac{\mathrm{c}+1-\mathrm{c}}{1-\mathrm{c}}=4$

⇒$\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1$

Q16. If a2 = b + c, b2 = c + a, c2 = a + b, then the value of

$\frac{1}{1+\mathrm{a}}+\frac{1}{1+\mathrm{b}}+\frac{1}{1+\mathrm{c}}$

(a) abc

(b) a2b2c2

(c) 1

(d) 0

Answer: (c) 1

Solution: $\frac{1}{1+\mathrm{a}}=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{a}^2}=\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$

Similarly, $\frac{1}{1+\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$

Similarly, $\frac{1}{1+\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$

∴$\frac{1}{1+\mathrm{a}}+\frac{1}{1+\mathrm{b}}+\frac{1}{1+\mathrm{c}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}=1$

Q17. If a + b + c = 0, then find the value of

$\frac{\mathrm{a}^2}{\mathrm{a}^2-\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{b}^2+\mathrm{ca}}+\frac{\mathrm{c}^2}{\mathrm{c}^2-\mathrm{ab}}$

(a) 0

(b) 1

(c) 6

(d) None of these

Answer: (d) None of these

Solution: Given, a + b + c = 0 ⇒ a = –(b+c) ⇒ a2 = (b+c)2

Now, $\frac{\mathrm{a}^2}{\mathrm{a}^2-\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{b}^2-+\mathrm{ca}}+\frac{\mathrm{c}^2}{\mathrm{c}^2-\mathrm{ab}}$

$=\,\,\frac{\left( \mathrm{b}+\mathrm{c} \right) ^2}{\left( \mathrm{b}+\mathrm{c} \right) ^2-\mathrm{bc}}+\frac{\mathrm{b}^2}{\mathrm{b}^2+\mathrm{c}\left( \mathrm{b}+\mathrm{c} \right)}+\frac{\mathrm{c}^2}{\mathrm{c}^2+\mathrm{b}\left( \mathrm{b}+\mathrm{c} \right)}$

$=\frac{\mathrm{b}^2+2\mathrm{bc}+\mathrm{c}^2}{\mathrm{b}^2+\mathrm{bc}+\mathrm{c}^2}+\frac{\mathrm{b}^2}{\mathrm{b}^2+\mathrm{c}^2+\mathrm{bc}}+\frac{\mathrm{c}^2}{\mathrm{b}^2+\mathrm{c}^2+\mathrm{bc}}$

$=\frac{2\left( \mathrm{b}^2+\mathrm{bc}+\mathrm{c}^2 \right)}{\mathrm{b}^2+\mathrm{bc}+\mathrm{c}^2}=2$

Q18. If a + b + c = 3, a2 + b2 + c2 = 6 and$\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}=1$, where a, b, c are all non-zero, then ‘abc’ is equal to

(a) $\frac{1}{2}$

(b) $\frac{1}{4}$

(c) $\frac{2}{3}$

(d) $\frac{3}{2}$

Answer: (d)$\frac{3}{2}$

Solution: a + b + c = 3

⇒ a2 + b2 + c2 +2ab +2bc +2ca = 9 (squaring both side)

⇒ ab + bc + ca = $\frac{3}{2}$

Now, $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}=1$

⇒ $\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{\mathrm{abc}}=1$

⇒ abc = ab + bc + ca = $\frac{3}{2}$

Q19. If x2 + y2 + z2 + 2 = 2(y – x), then value of x3 + y3 + z3 is equal to

(a) 0

(b) 1

(c) 2

(d) 3

Answer: (a) 0

Solution: x2 + y2 + z2 + 2 = 2(y – x)

⇒ x2 + 2x + y2 – 2y + z2 + 2 = 0

⇒ (x2 + 2x + 1) + (y2 – 2y + 1) + z2 = 0

⇒ (x + 1)2 + (y – 1)2 + z2 = 0

⇒ x + 1 = 0 ⇒ x = – 1; y – 1 = 0 ⇒ y = 1; z = 0

∴ x3 + y3 + z3 = – 1 + 1 + 0 = 0

Q20. Q. If   $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}=119$ and x > 1 then the value of $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$

(a) 54

(b) 18

(c) 72

(d) 36

Answer: (d) 36

Solution: $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}=119$

⇒$\mathrm{x}^4+\frac{1}{\mathrm{x}^4}+2.\mathrm{x}^2.\frac{1}{\mathrm{x}^2}=121$

⇒ $\left( \mathrm{x}^2+\frac{1}{\mathrm{x}^2} \right) ^2=11^2$

⇒ $\left( \mathrm{x}^2+\frac{1}{\mathrm{x}^2} \right) =11$ ( x > 1)

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}-2.\mathrm{x}.\frac{1}{\mathrm{x}}=9$

⇒$\left( \mathrm{x}-\frac{1}{\mathrm{x}} \right) ^2=3^2$

⇒$\mathrm{x}-\frac{1}{\mathrm{x}}=3$ ( X > 1)

Using Trick ⇒$\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,3^3+3\times 3=36$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}$ = n3 – 3.n and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$ = n3 + 3.n

 

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